12. Introduction to Three dimensional geometry (Question Bank)
I .ONE MARK QUESTIONS WITH ANSWERS:
1. Name the octant in which the point (4, −2, −5) lie?
Ans:(4, −2, −5) lies in 8th octant 2. Name the octant in which the point (−2, −4, −7) lie?
Ans:(−2, −4, −7) lies in 7th octant .
3. Name the octant in which the point (−3,1, −2) lie.
Ans: (−3,1, −2) lies in 6th octant .
4. Find the distance between the pair of points (−3,7,2) & (2,4, −1).
Ans: Required distance= (−3 − 2) + (7 − 4) + (2 + 1)
= √25 + 9 + 9 = √43.
5. Find the distance between pair of points (2, −1,3) (−2,1,3).
Ans: Required distance = (2 + 2) + (−1 − 1) + (3 − 3)
= √16 + 4 = √20.
6. Find the coordinates of the foot of the perpendicular of the point (2,4,5) in the XZ plane.
Ans: (2,0,5).
7. Write the coordinate of any point in the XY plane.
Ans:(, , 0).
8. Find the equation of YZ plane.
Ans: = 0.
9. Find the equation of ZX plane.
Ans:
= 0.
10. Find the reflexion (Image) of the point !– , #, $%& 'ℎ) *+).
Ans: (, #, $).
1
11. Find the reflexion (Image) of the point (2,3, −4)& 'ℎ) *+).
Ans: (2,3,4).
12. Find the coordinate of the centroid of the triangle formed by the points
(2, −5,1), (3,2, −6) & (1, −3, −1)
Ans: G≡ -
./ 0.1 0.2 4/ 041 042 5/ 051 052
,
,
6
3
3
3
≡ (2, −2, −2).
13. Find the coordinates of midpoint of line joining the points 7 ≡ (−5, −3,7) & 8 ≡ (−7,3,5)
Ans: ≡ -
./ 0.1 4/ 041 5/ 051
,
,
6 ≡ (−6,0,6).
II. TWO MARKS QUESTIONS WITH ANSWERS:
1. Show that the points (−2,3,5)9(1,2,3) :(7,0, −1) are collinear.
Ans: 9 = (−2 − 1) + (3 − 2) + (5 − 3) = √9 + 1 + 4 = √14.
9: = (1 − 7) + (2 − 0) + (3 + 1) = √36 + 4 + 16=√56 = 2√14
: = (−2 − 7) + (3 − 0) + (5 + 1) = √81 + 9 + 36 = √126 = 3√14
∴ 9 + 9: = :.
∴ =ℎ) *>&'? , 9 : are collinear.
2. Show that the points (3, −2,4), (1,1,1) (−1,4, −2) are collinear.
Ans: Let ≡ (3, −2,4) 9 ≡ (1,1,1) : ≡ (−1,4, −2)
∴ 9 = (3 − 1) + (−2 − 1) + (4 − 1) = √4 + 9 + 9 = √22.
9: = (1 + 1) + (1 − 4) + (1 + 2) = √4 + 9 + 9 = √22
: = (3 + 1) + (−2 − 4) + (4 + 2) = √16 + 36 + 36 = 2√22
∴ 9 + 9: = :
∴ The points , 9, : are collinear.
3. Find the coordinates of the point which divides the line segment joining the points
(1, −2,3) (3,4, −5) internally in the ratio 2: 3.
Ans: ≡ -
(3)03(A) (B)03(C) (CD)03(3)
, 03 , 03 6
03
E CA
6.
D
≡ -D , D ,
2
4. Find the coordinates of the point which divides the line segment joining the points
(1, −2,3) (3,4, −5) externally in the ratio 2: 3.
Ans: ≡ -
(3)03(A) (B)03(C) (CD)03(3)
C3
,
C3
,
3
AB CAE
6 = -CA , CA ,
C3
CA
6 ≡ (−3, −14,19).
5. Find the coordinates of the point which divides the line segment joining the points
(−2,3,5) (1, −4,6) internally in the ratio 2: 3.
Ans: Let 7 = (−2,3,5) 8 = (1, −4,6)
Let P divides AB internally in the ratio 2:3
≡-
(A)03(C) (CB)03(3) (F)03(D)
,
,
6
03
03
03
,∴ ≡ -
CB A G
, , 6.
D D D
6. Find the coordinates of the point which divides the line segment joining the points
(−2,3,5) (1, −4,6) externally in the ratio 2: 3.
Ans: Let 7 = (−2,3,5) 8 = (1, −4,6)
Let P divides AB externally in the ratio 2:3
≡-
(A)C3(C) (CB)C3(3) (F)C3(D)
C3
,
C3
,
C3
H
6 , ∴ ≡ -CA ,
CAG C3
CA
,
CA
6 ≡ (−8,17,3)
7. Find the ratio in which the YZ plane divides the line segment formed by joining the points
(−2,4,7) (3, −5,8) .
Ans: Let YZ plane divide line segment joining the points 7 ≡ (−2,4,7) 8 ≡ (3, −5,8) '
≡ (, , I)& 'ℎ) J'&> K: 1.
∴≡-
L(3)0A(C) L(CD)0A(B) L(H)0A(G)
,
,
6
L0A
L0A
L0A
∴≡-
3LC CDL0B HL0G
,
,
6
L0A
L0A
L0A
Given that the point ‘P’ lies on YZ plane⟹ $>>J&') >N = 0 ⟹
3LC
L0A
Therefore ratio is 2:3.
8. The centroid of a ∆78P is the point (1, 1, 1).If the coordinates of A and B are
(3, −5,7) (−1,7, −6)
respectively. Find the coordinates of the point ‘C’.
Ans: Let P ≡ (, , I); 7 ≡ (3, −5,7) 8 ≡ (−1,7, −6)
./ 0.1 0.2 4/ 041 042 5/ 051 052
,
,
6
3
3
3
G≡ -
3
3
=0⟹K= .
⟹ (1,1,1) ≡ -
3CA0. CD0G04 GCF05
⟹ (1,1,1) ≡ -
0. 04 A05
,
3
,
3
3
,
3
,
3
3
6
6
⟹ (, , I) ≡ (1,1,2)
⟹ P ≡ (1,1,2).
9. The vertices of a triangle are (2, 4, 6) (0,-2,-5).If origin is the centroid of the triangle find the third
vertex?
Ans: Let 7 ≡ (2,4,6) 8 ≡ (0, −2, −5) P ≡ (, , I) be the vertices of ∆78P
./ 0.1 0.2 4/ 041 042 5/ 051 052
G≡ -
3
,
,
3
3
6
⟹ (0,0,0) ≡ -
0R0. BC04 FCD05
⟹ (0,0,0) ≡ -
0. 04 A05
,
3
3
,
3
,
3
,
3
3
6
6
⟹ (, , I) ≡ (−2, −2, −1)
⟹ P ≡ (−2, −2, −1).
10. If the distance between the points (k,-8, 4) and (3,-5, 4) is 5. Then find k.
Ans: Let A≡ (K, −8,4) and B≡ (3, −5,4)
Given AB= 5 ⟹ (K − 3) + (−8 + 5) + (4 − 4) = 5
⟹ √K + 9 − 6K + 9 = 5
⟹ K − 6K − 7 = 0
⟹ K − 7K + K − 7 = 0
⟹ K(K − 7) + 1(K − 7) = 0
⟹ K = −1 >J K = 7.
11. Three consecutive vertices of a parallelogram ABCD are 7 ≡ (3, −1,2) 8 ≡ (1,2, −4)
P ≡ (−1,1,2). Find the coordinates of fourth vertex D.
A
Ans: Let S ≡ (, , I) be fourth vertex of parallelogram
B
4
D
C
⟹ Midpoint of AC = Midpoint of BD.
⟹-
3CA CA0A 0
,
,
6
⟹ (1,0,2) = ⟹ = 1,
A0. 04 CB05
,
,
6
=-
A0. 04 CB05
,
,
6
= −2, I = 8
∴ S ≡ (1, −2,8)
12. If the origin is the centroid of the triangle PQR with the vertices
≡ (2, 2,6), 9 ≡ (−4,3#, −10) : ≡ (8,14,2$) then find the values of a, b, c.
Ans: Given origin = Centroid of a triangle PQR.
TCB0H 03U0AB FCAR0V
,
,
6
3
3
3
⟹ (0,0,0) = -
T0B 3U0AF VCB
,
,
6
3
3
3
⟹ (0,0,0) = -
⟹ = −2, # = −
AF
,$
3
= 2.
13. Centroid of a triangle with vertices 7 ≡ (, 1,3), 8 ≡ (−2, #, −5) P ≡ (4,7, $) is the origin.
Find the values of a, b, c.
Ans: Given origin = Centroid of a triangle ABC.
TC0B A0U0G 3CD0V
, 3 , 3 6
3
⟹ (0,0,0) = -
⟹ = −2, # = −8, $ = 2.
III. FIVE MARKS QUESTIONS WITH ANSWERS:
1. Show that the points W ≡ (X, Y, Z) [ ≡ (−X, −Y, −X) \ ≡ (Y, Z, Y) ]^_ ` ≡ (a, b, c) are the
vertices of parallelogram ABCD but it is not a rectangle.
Ans: To show ABCD is a parallelogram we need to show that two opposite sides are equal.
W[ = (X + X)Y + (Y + Y)Y + (Z + X)Y = √a + Xc + Xc = √Zc = c.
[\ = (−X − Y)Y + (−Y − Z)Y + (−X − Y)Y = √d + Ye + d = √aZ
\` = (Y − a)Y + (Z − b)Y + (Y − c)Y = √a + Xc + Xc = √Zc = c
`W = (X − a)Y + (Y − b)Y + (Z − c)Y = √d + Ye + d = √aZ
5
W\ = (X − Y)Y + (Y − Z)Y + (Z − Y)Y = √Z
[` = (−X − a)Y + (−Y − b)Y + (−X − c)Y = √Ye + fX + ad = √Xee.
AC≠BD.Diagonals are not equal.∴ ABCD is not a rectangle.
2. Show that the points (-1,2,1) (1,-2,5) (4,-7,8) and (2,-3,4) are the vertices of the parallelogram.
Ans: Let W ≡ (−X, Y, X) [ ≡ (X, −Y, e) \ ≡ (a, −b, f)
]^_ ` ≡ (Y, −Z, a)
hi = (−X − X)Y + (Y + Y)Y + (X − e)Y = √a + Xc + Xc = √Zc = c.
[\ = (X − a)Y + (−Y + b)Y + (e + f)Y = √d + Ye + d = √aZ
\` = (a − Y)Y + (−b + Z)Y + (f − a)Y = √a + Xc + Xc = √Zc = c
`W = (−X − Y)Y + (Y + Z)Y + (X − a)Y = √d + Ye + d = √aZ
∴ W[ = \` ]^_ [\ = `\
W\ = (−1 − 4)Y + (2 + 7)Y + (1 − 8)Y = √25 + 81 + 49 = √155
[` = (1 − 2)Y + (−2 + 3)Y + (5 − 4)Y = √1 + 1 + 1 = √3.
AC≠BD.Diagonals are not equal.∴ ABCD is a Parallelogram.
3. Show that the points (j, b, −Xj), (X, c, −c) ]^_(a, d, −c) are the vertices of an isosceles right
angled triangle.
Ans: Let W ≡ (j, b, −Xj), [ ≡ (X, c, −c) ]^_ \ ≡ (a, d, −c)
Now W[ = (j − X)Y + (b − c)Y + (−Xj + c)Y = √X + X + Xc = √Xf
[\ = (X − a)Y + (c − d)Y + (−c + c)Y = √d + d + j = √Xf
W\ = (j − a)Y + (b − d)Y + (−Xj + c)Y = √Xc + a + Xc = √Zc
Clearly W[ = [\ ]^_ W[Y + [\Y = W\Y
∴ The given points are the vertices of an isosceles right angled triangle.
4. Show that the points (j, b, Xj), (−X, c, c) ]^_(−a, d, c) are the vertices of an isosceles right angled
triangle.
Ans: Let W ≡ (j, b, Xj), [ ≡ (−X, c, c) ]^_ \ ≡ (−a, d, c)
Now W[ = (j + X)Y + (b − c)Y + (Xj − c)Y = √X + X + Xc = √Xf
6
[\ = (−X + a)Y + (c − d)Y + (c − c)Y = √d + d + j = √Xf
W\ = (j + a)Y + (b − d)Y + (Xj − c)Y = √Xc + a + Xc = √Zc
Clearly W[ = [\ ]^_ W[Y + [\Y = W\Y
∴ The given points are the vertices of an isosceles right angled triangle.
5. Find the lengths of the medians of the triangle with the vertices A(0,0,6) B(0,4,0) and C (6,0,0).
Ans: Let ` =Midpoint of BC
∴`=-
j0c a0j j0j
Y
,
Y
,
Y
6 = (Z, Y, j)
∴Length of median W` = (j − Z)Y + (j − Y)Y + (c − j)Y = √d + a + Zc = √ad = b
Let k =Midpoint of CA
∴k=-
c0j j0j j0c
Y
,
Y
,
Y
6 = (Z, j, Z)
∴Length of median [k = (j − Z)Y + (a − j)Y + (j − Z)Y = √d + Xc + d = √Za
Let l =Midpoint of AB
∴l=-
j0j j0a c0j
Y
,
Y
,
Y
6 = (j, Y, Z)
∴Length of median\l = (c − j)Y + (j − Y)Y + (j − Z)Y = √Zc + a + d = √ad = b.
6. Show that the points (j, a, X), (Y, Z, −X), (a, e, j) ]^_ (Y, c, Y) are the vertices of the square.
Ans. Let W ≡ (j, a, X), [ ≡ (Y, Z, −X), \ ≡ (a, e, j) ]^_ ` ≡ (Y, c, Y)
Now W[ = (j − Y)Y + (a − Z)Y + (X + X)Y = √a + X + a = √d = Z
[\ = (Y − a)Y + (Z − e)Y + (−X − j)Y = √a + a + X = √d = Z
\` = (a − Y)Y + (e − c)Y + (j − Y)Y = √a + X + a = √d = Z
`W = (Y − j)Y + (c − a)Y + (Y − X)Y = √a + a + X = √d = Z
∴ W[ = [\ = \` = `W = Z
∴ All the sides are equal.
Wmno W\ = (j − a)Y + (a − e)Y + (X − j)Y = √Xc + X + X = √Xf
[` = (Y − Y)Y + (Z − c)Y + (−X − Y)Y = √j + d + d = √Xf
7
∴ W\ = [` ∴ The diagonals are equal. ∴ W[\` is a square.
7. Show that the points W ≡ (X, Z, a), [ ≡ (−X, c, Xj), \ ≡ (−b, a, b) ]^_ ` ≡ (−e, X, X) are the
vertices of rhombus.
Ans: W[ = (X + X)Y + (Z − c)Y + (a − Xj)Y = √a + d + Zc = √ad = b
[\ = (−X + b)Y + (c − a)Y + (Xj − b)Y = √Zc + a + d = √ad = b
\` = (−b + e)Y + (a − X)Y + (b − X)Y = √a + d + Zc = √ad = b
`W = (−e − X)Y + (X − Z)Y + (X − a)Y = √Zc + a + d = √ad = b
∴ W[ = [\ = \` = `W = b
∴ All the sides are equal.
Wmno W\ = (X + b)Y + (Z − a)Y + (a − b)Y = √ca + X + d = √ba
[` = (−X + e)Y + (c − X)Y + (Xj − X)Y = √Xc + Ye + fX = √XYY
∴ W\ ≠ [` ∴ The diagonals are not equal. ∴ The given points form a rhombus.
8. Derive an expression for the coordinates of a point that divides the line joining the points.
W ≡ (pX , qX , rX ) ]^_ [ ≡ (pY , qY , rY ) Internally in the ratio m: n Hence find the coordinates of the
midpoint of AB.where W ≡ (X, Y, Z) ]^_ [ ≡ (e, c, b)
Z
B
Ans: Let the s ≡ (p, q, r) devide the line segment
C
P
Joining the two points W ≡ (pX , qX , rX ) and
D
A
[ ≡ (pY , qY , rY ) Internally in the ratio m: n.
Draw AL, BM and PN ⊥ to XY plane.
Clearly AL, BM and PN lie on the plane containing AB
But ⊥ to XY plane.
O
The feet of perpendiculars L, M and N are collinear.
Y
Through P draw a line parallel to the line LM
M
Meets AL produced and BM at C and D.
L
Clearly ∆ PAC and ∆PDB are equiangular
And hence similar.
8
X
N
∴
uv
=
uw
vx
(1)
wy
Now 7P = Pz − 7z = { − 7z = I − IA
8S = 8| − S| = 8| − { = I − I
Now (1) becomes
}
~
=
5C5/
51C5
⟹ (I − IA ) = (I − I)
⟹ I − IA = I − I
⟹ ( + )I = I + IA
⟹I=
}51 0~5/
}0~
.
Similarly we can prove that
=
}.1 0~./
}0~
∴≡-
=
,
}41 0~4/
}0~
}.1 0~./ }41 0~4/ }51 0~5/
,
}0~
,
}0~
}0~
6.
A0D 0F 30G
Midpoint of 78 ≡ -
,
,
6
9. Find the distance between two points in a three dimensional plane and hence find the
distance between the points ≡ (−2,3,5) 9 ≡ (1,2,3).
Ans: Let‘O’be the origin. OX, OY, OZ be the axes
Let 7 ≡ (A ,
A , IA ) 8 ≡ ( ,
B
Z
, I )
A
Be any two points in the 3D-plane.
L
Draw AM and BN ⊥ to XY-plane and
AL⊥ to BN.
O
Clearly |{ = ( − A ) + (
A)
−
Now 7z = |{ = ( − A ) + (
−
Y
A)
M
X
9
N
Also8z = 8{ − z{ = 8{ − 7| = I − IA .
From the right angled ∆ALB we have,
78 = 7z + 8z = ( − A ) + (
∴ 78 = ( − A ) + (
−
A)
A)
−
+ (I − IA )
+ (I − IA )
Also,
Distance 9 = (−2 − 1) + (3 − 2) + (5 − 3)
= √9 + 1 + 4 = √14.
10. The midpoints of the sides of a triangle are (5,7,11), (0,8,5) (2,3, −1) Find its vertices.
Ans: Let S = |&*>&' >N 78 = (5,7,11)
C(3 ,
€ = |&*>&' >N 8P = (0,8,5)
 = |&*>&' >N 7P = (2,3, −1)
Where 7 ≡ (A ,
A , IA ), 8
≡ ( ,
, I )
E
F
P ≡ (3 ,
3 , I3 )
3 , I3 )
Are the vertices of ∆78P.
./ 0.1 4/ 041 5/ 051
Now (5,7,11) = -
,
,
(1)
⟹ A + = 10
A
+
(2)
= 14
(3)
IA + I = 22
(2,3, −1) = -
.1 0.2 41 042 51 052
A(A ,
6
,
,
6
(4)
⟹ + 3 = 4
=6
(5)
I + I3 = −2
(6)
+
3
./ 0.2 4/ 042 5/ 052
Also,(0,8,5) = -
,
⟹ A + 3 = 0
10
,
6
(7)
A , IA )
D
B( ,
, I )
+
A
3
(8)
= 16
(9)
IA + I3 = 10
Adding (1) + (4) + (7)
⟹ 2A + 2 + 23 = 14
⟹ A + + 3 = 7
⟹ 10 + 3 = 7 ⟹ 3 = −3
From (7)
A + 3 = 0 ⟹ A = 3
From (1) A + = 10 ⟹ = 7
(2) + (5) + (8) gives,
⟹ 2
⟹
A
+2
3
+2
3
+
+
⟹ 14 +
3
= 18 ⟹
A
J> (8),
From (2),
A
A
+
= 36
= 18
+
3
3
=4
= 16 ⟹
= 14 ⟹
A
= 12
=2
(3) + (6) + (9) gives,
⟹ 2IA + 2I + 2I3 = 30
⟹ IA + I + I3 = 15
⟹ 10 + I3 = 15 ⟹ I3 = 5
J> (9),
IA + I3 = 10 ⟹ IA = 5
From (3), IA + I = 22 ⟹ I = 17
7 ≡ (A ,
A , IA )
≡ (3,12,5)
8 ≡ ( ,
, I )
≡ (7,2,17)
P ≡ (3 ,
3 , I3 )
≡ (−3,4,5) are the required points.
11